What is the foci x^2-4x+8y-4=0 ?

The foci x^2-4x+8y-4=0 is (2,-1)

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foci x^2-4x+8y-4=0

foci

x^{2} - 4 x + 8 y - 4 = 0

(2, - 1)

Solution steps

Rewrite

x^{2} - 4 x + 8 y - 4 = 0

in the standard form

4 (- 2) (y - 1) = (x - 2)^{2}

(h, k) = (2, 1), p = - 2

Parabola is symmetric around the y-axis and so the focus lies a distance

p

from the center

(2, 1)

along the y-axis

(2, 1 + p)

= (2, 1 + (- 2))

Refine

(2, - 1)

- 2 x^{2} + 4 x + 3

- 2 x^{2} + 4 x + 7

- 2 x^{2} + 4 x - 7

6 y = x^{2} - 8 x + 14

(x - 1)^{2} = - 8 (y 2)